Aptitude Discussion

Q. |
Walking at the rate of 4 kmph a man cover certain distance in 2 hr 45 min. Running at a speed of 16.5 kmph the man will cover the same distance in. |

✖ A. |
12 min |

✖ B. |
25 min |

✔ C. |
40 min |

✖ D. |
60 min |

**Solution:**

Option(**C**) is correct

Distance = Speed × time

Here time = 2hr 45 min = \(\dfrac{11}{4}\) hr

Distance = \(4\times \dfrac{11}{4}=11\) km

New Speed =16.5 kmph

Therefore time = \(\dfrac{D}{S}=\dfrac{11}{16.5}=\textbf{40 min}\)

**Edit:** For an alternative solution, check **Debanjan's** comment.

**Poonam**

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**Vaishali**

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in that last step how did u get the ans 11/16.5 = 40 ??????

You are finding the time taken in minutes...11/16.5= 0.66 and 0.66*60=40 min

**Arpan Bansal**

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Alternate solution can be:

V1/V2=T2/TI

4 Kmph/ 16.5 Kmph = T2 / 165 Min 2hr 45min = 165 min

After solving

T2 = 40 min

**Akshay**

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No need to find the distance ,as speed ti increased approx 4 time so the time is reduced by 1/4 so simply 4/14.5 * 165 min= 40 min .I think simple approach

**Unknown**

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distance cover = 4*165=x*165/10

x=40 min

**NEHA**

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what is V1, V2? AND HOW IT IS 4/33/2?

**Balaji**

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in above explanation how it will come 2hr 45min=11/4hr ?

2hr 45min = 2hr +$\dfrac{3}{4}$ hr

$= 2=\dfrac{3}{4}$ hr

$=\dfrac{11}{4}$ hr.

Hope that helps.

**Debanjan**

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It is much easier to use the formula $\dfrac{V_1}{V_2}=\dfrac{T_2}{T_1}$

or, $\dfrac{4}{33/2}=\dfrac{T_2}{11/4}$

or $T_2 = \dfrac{2}{3}$ hour

$= \dfrac{2}{3}\times 60\text{ min}$

$=\textbf{40min}$

debanjan solved the question but how 2/3 came ? can any 1 explain pls